Let $h$ be a vector-valued function defined by $h(t)=(\log(10t),\sin(t))$. Find $h'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\dfrac{1}{t\ln(10)},\cos(t)\right)$ (Choice B) B $\left(\dfrac{1}{10t},\cos(t)\right)$ (Choice C) C $\left(\dfrac{1}{t},-\cos(t)\right)$ (Choice D) D $\dfrac{1}{10t}-\cos(t)$
Answer: $h$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $h(t)=(\log(10t),\sin(t))$. Let's differentiate the first expression: $\dfrac{d}{dt}[\log(10t)]=\dfrac{1}{t\ln(10)}$ Let's differentiate the second expression: $\dfrac{d}{dt}[\sin(t)]=\cos(t)$ Now let's put everything together: $\begin{aligned} h'(t)&=\left(\dfrac{d}{dt}\left[\log(10t)\right],\dfrac{d}{dt}[\sin(t)]\right) \\\\ &=\left(\dfrac{1}{t\ln(10)},\cos(t)\right) \end{aligned}$ In conclusion, $h'(t)=\left(\dfrac{1}{t\ln(10)},\cos(t)\right)$.